Question

The pK_b, values for the dibasic base B are pK_b1 = 2.10 and pKb2 = 7.75 Calculate the pH at each of the following points In the titration of 50 0 mL of a 0 50 M B(ag) with 0.50 M HCI(ag). before addition of any HCI after addition of 25 0 mL of HCI after addition of 50 0 mL of HCI after addition of 75.0 ml. of HCI after addition of 100 0 mL of HCI

Answer 1

a) First dissociation:

B^- + H2O ---> BH⁺ + OH^-

Kb1 = ([BH+]*[OH-]) / [B-]

Remeber Kb1 = 10^-pKb1 = 7.94*10^-3

7.94*10^-3 = (x*x) / (0.5 M - x)

x = 0.059 M = [OH-] = [BH⁺]

Second dissociation:

BH⁺ + H2O --> BH₂⁺² + OH^-

Kb2 = [BH₂⁺²] * [OH^-] / [BH⁺]

Kb2 = 10^-pKb2

3.2*10^-8 = (x)*(0.059+x) / (0.059-x)

x = 0.059

[OH-] = 0.059 + x = 0.059 + 0.059 = 0.118

pOH = -log([OH-]) = 0.93

pH = 14 - 0.93 = 13.1

b) Volumen of HCl at first equivalence point :

V = 0.5 M * 50 ml / 0.5 M = 50 ml

So in 25 ml you're half way to the equivalence point:

pH = pKa1 = 14 - pKb1 = 11.9

c) At first equivalence point:

pH = 1/2 ( pKa1 + pKa2) = 9.2

d) Volume at second equivalence point = 2* 50 ml = 100 ml

When 75 ml has been added, you're on the half way to the second equivalence point, so:

pH = pKa2 = 14 - pKb2 = 6.5

e) At second equivalence point

BH₂⁺² + H2O --> BH⁺ + H3O⁺

Ka2 = [BH+] * [H3O+] / [BH₂+2]

Where, [BH₂+2] = (50 ml * 0.5 M) / (50 ml + 100 ml) = 0.17 M

and Ka2 = 10^-6.5

10^-6.5 = (x*x) / (0.17 -x)

x = 0.00023 = [H3O+]

pH = -log([H3O+]) = 3.64

Answer 2

a) Before addition of any HCl

First dissociation:

                      B   +   H2O <----------------------> BH+   + OH-

Kb1 = [BH+][OH-]/[B]

7.94 x 10^-3 = x^2 / 0.50 -x

On solving,

x = 0.059 M

[OH-] = 0.059 M

Second dissociation:

BH+ + H2O ----> BH₂⁺² + OH-

Kb2 = [BH₂⁺²][OH-]/[BH+]

1.78*10^-8 = (x) * (0.059 +x) / (0.059 -x)

x = 0.059

[OH-] = 0.059 + 0.059 = 0.118+

pOH = -log([OH-])= 0.93

pH = 14 - 0.93 = 13.07

(b) After additon of 25 mL HCl

Volume of first equivalence point = (50 ml * 0.5M) / 0.5 M = 50 ml

It is first half equilivalence point.

at this point :

pH = pKa1 = 14 - pKb1 = 11.9

c) After additon of 50 mL HCl

at first equivalence point

pH = 1/2 ( pKa1 + pKa2) = 1/2 (11.9 + 6.25) = 9.08

(d) After additon of 75 mL HCl

it is second half equivalence point

pOH = pKb2

pOH = 7.75

pH = 6.25

e) After additon of 100 mL HCl

Molarity of BH₂⁺² = (50 ml * 0.5 M) / (50 ml +100 ml) = 0.167 M

BH2+2 -----------------> BH+     +    H+

0.167                        0 0

0.167 -x                     x    x

Ka2 = x^2 / 0.167 -x

pKa2 = 14 - 7.75 = 6.25

Ka2 = 10^-pKa2

Ka2 = 5.62*10^-7

5.62*10^-7 = x^2 / 0.167 -x

x = 0.00031 = [H+]

pH = -log [H+]

pH = 3.51